Problem 24 - Advanced Engineering Mathematics

Question

For \(y^{\prime}+\frac{2}{x^{2}-1} y=\frac{x+1}{x-1}\) an integrating factor is \(e^{\int\left[2 /\left(x^{2}-1\right)\right] d x}=\frac{x-1}{x+1}\) so that \(\frac{d}{d x}\left[\frac{x-1}{x+1} y\right]=1\) and \((x-1) y=x(x+1)+c(x+1)\) for \(-1< x<1\).

Step-by-Step Solution

Answer

The solution is \( (x-1)y = x(x+1) + C(x+1) \), for \(-1 < x < 1\).

Step 1: Understanding the Given Equation

The differential equation given is \( y^{\prime}+\frac{2}{x^{2}-1} y=\frac{x+1}{x-1} \). We need to solve this first-order linear differential equation.

Step 2: Identifying the Integrating Factor

An integrating factor is required to simplify the equation. The integrating factor given is \( e^{\int \left[\frac{2}{x^{2}-1} \right] dx} = \frac{x-1}{x+1} \). This factor is derived from integrating \( \frac{2}{x^{2}-1} \).

Step 3: Applying the Integrating Factor

Multiply the entire differential equation by the integrating factor \( \frac{x-1}{x+1} \) to get the equation \( \frac{d}{dx} \left[\frac{x-1}{x+1} y \right] = 1 \). This simplifies the equation to a form where the left side is the derivative of a single function.

Step 4: Integrating Both Sides

Integrate both sides with respect to \( x \). The left side becomes \( \frac{x-1}{x+1} y \), and the integral of the right side (1) with respect to \( x \) is \( x + C \), where \( C \) is the constant of integration.

Step 5: Solve for \( y \)

From the integration, we have \( \frac{x-1}{x+1} y = x + C \). Solve for \( y \) by multiplying both sides by \( \frac{x+1}{x-1} \) to get \( y = (x + C) \cdot \frac{x+1}{x-1} \).

Step 6: Simplifying the Solution

Distribute to simplify \( y = x \left(\frac{x+1}{x-1}\right) + C \left(\frac{x+1}{x-1}\right) \), leading to \( y = \frac{x(x+1) + C(x+1)}{x-1} \). Combine terms to see the general solution as given: \( (x-1) y = x(x+1) + C(x+1) \) for \(-1< x<1\).

Key Concepts

Integrating Factor

An integrating factor is a powerful tool used to solve first-order linear differential equations. It transforms the equation into a simpler form that is easier to integrate. To find an integrating factor, you start by identifying the coefficient of the function \( y \) in the differential equation:

• Initial equation: \( y^{\prime} + P(x)y = Q(x) \)
• Integrating factor: \( \mu(x) = e^{\int P(x) \, dx} \)

In our case, the differential equation is \( y^{\prime} + \frac{2}{x^2 - 1} y = \frac{x+1}{x-1} \). Thus, the integrating factor is derived from the term \( \frac{2}{x^2 - 1} \). When \( \mu(x) \) is applied, it makes the left-hand side of the equation:

• \( \frac{d}{dx} [\mu(x) y] = \mu(x) Q(x) \)

This enables the differential equation to convert into an exact equation that you can easily integrate. By applying the integrating factor \( \mu(x) = \frac{x-1}{x+1} \), the equation transforms into \( \frac{d}{dx} \left[ \frac{x-1}{x+1} y \right] = 1 \). This simplified form allows for straightforward integration of both sides.

First-Order Linear Differential Equations

First-order linear differential equations involve derivatives of the first order. They are characterized by their straightforward structure, allowing them to be solved using methods such as integrating factors. These equations generally look like:

• \( y^{\prime} + P(x)y = Q(x) \)

Here, \( P(x) \) is a function of \( x \), and similarly, \( Q(x) \) can be any function dependent on \( x \). The procedure to solve involves:

• Calculating the integrating factor using \( P(x) \).
• Multiplying throughout by this integrating factor to make the equation exact.
• Solving the resulted exact differential equation through integration.

In our example involving the equation \( y^{\prime} + \frac{2}{x^2 - 1} y = \frac{x+1}{x-1} \), the goal was to manipulate the equation such that it could be integrated easily. The equation after integrating factor application becomes \( \frac{d}{dx} \left[ \frac{x-1}{x+1} y \right] \), equating to a function of \( x \) that is integrable: 1. This systematic approach not only simplifies the solving process but also ensures that the solution can be found using direct integration.

Constant of Integration

The concept of the constant of integration arises when you integrate a differential equation. During the integration process, especially when solving first-order linear differential equations, you will often encounter indefinite integrals.

• An indefinite integral is characterized by its general form: \( \int f(x) \, dx = F(x) + C \).
• Here \( C \) represents the constant of integration.

This constant accounts for all possible vertical shifts in the solution, given that there are many functions whose derivatives can be the same. When integrating both sides in the stepped solution of the differential equation, such as in:

• \( \frac{d}{dx} \left[ \frac{x-1}{x+1} y \right] = 1 \)
• Results in: \( \frac{x-1}{x+1} y = x + C \)

The \( C \) encompasses different initial conditions that the solution might satisfy. Solving for \( y \) incorporates this constant, reflecting in the final solution \( (x - 1)y = x(x + 1) + C(x + 1) \). This inclusion is crucial as it ensures the solution is adaptable to particular sets of initial conditions.